A chemistry course to cover selected topics covered in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society.
Prerequisites: Students should have a background in basic chemistry including nomenclature, reactions, stoichiometry, molarity and thermochemistry.

Taught By

Dr. Allison Soult

Dr. Kim Woodrum

Transcript

Well, we have to introduce to you the concept of a common ion effects on molar solubility. We saw the common ion effect in the world of the buffer. We are going to see it here again and see how adding a common ion affects the solubility. The solubility we are going to learn in what way it effects it and we are going to be able to calculate the solubility after a common ion has been added. So lets consider this equilibrium. If we were to take the silver chloride and add it to a solution If we were to take the silver chloride and add it to a solution and have an equilibrium be established and then go in and add some sodium chloride to that solution we would be increasing the amount of Cl- that is in that solution. Le Chatelier's principle tells us way this equilibrium will shift. Select your answer. Well, if you said it would shift it to the left then you are correct. Le Chatelier's principle says that if add a substance on one side it will shift away from that substance and try to obtain, re-establish equilibrium. Now if it is shifting it to the left what can we say about the solubility of silver chloride? Is it increasing the amount that will dissolve or is it decreasing the amount that dissolves? And I have it written the other way, decreasing and increasing. Well, if you are decreasing the solubility that is correct. For pushing this to the left there is more solid, less ions in solution that by definition is a decrease in the solubility. So the common ion effect of molar solubility is always the same. If you add a common ion to this solution it will always decrease the solubility of the salt. Now it is important for you to understand that it does not change the K_sp . K_sp is a constant that is the solubility product and it is a constant so that is not changing. solubility product and it is a constant so that is not changing. But the S the molar solubility is decreasing as you add the common ion. So lets see how you work through one of these problems. is decreasing as you add the common ion. So lets see how you work through one of these problems. I have magnesium fluoride and I want to know its molar solubility when I have added it to a solution that contains some sodium fluoride. So the first thing is if this is the salt that I am dissolving which one of these two ions is the common ion? Well, it is the F-. Now, when you work these problems students struggle with writing the reaction because they take these two substances they want to take this and this and react them together. That is not the reaction. The reaction is the same reaction we have been always writing. We are taking the insoluble salt and we are dissolving it into water, or we are dissolving it. So we have the ions it is aqueous, and we have 2 F- thats aqueous. That is the reaction we are using the same reaction we have always written. I will have an ICE table and this is where it is different. I will have an ICE table and this is where it is different. I am placing some of this insoluble salt into this beaker. I am placing some of the MgF_2 into this beaker. But this beaker is no longer just water. This beaker contains this sodium ions already and some F- ions because it is sodium fluoride. So the F- is a common ion and it is already present. So I am going to put a zero under magnesium because there is not already present some magnesium. But there is already present some F-. because there is not already present some magnesium. But there is already present some F-. Now this sodium fluoride when it dissolves But there is already present some F-. Now this sodium fluoride when it dissolves breaks apart 100% into Na+ and F-. So if the concentrations of NaF and F-. So if the concentrations of NaF is .10, that is the concentration of F-. Now a common mistake that I see students making is to try to incorporate this 2 somehow. Now a common mistake that I see students making is to try to incorporate this 2 somehow. It does not come into play in the initial line. It has nothing to do with the initial line. That too has to do with the reaction taking place and that is always in the change line. You are just plugging it in because you know right now before anything happens with the magnesium fluoride that there is already some fluoride present and the amount already present is point .10. So now our change line takes place, there is nothing different here. and the amount already present is point .10. So now our change line takes place, there is nothing different here. Some of the magnesium fluoride dissolves and that is the molar solubility for everyone one of those that dissolves you make one of these. and two of the fluoride. So now there is still a large excess, so there is some and two of the fluoride. So now there is still a large excess, so there is some still sitting on the bottom. I will have S here and I will have .10 2S here. As I write the expression, the K_sp expression I will write it first, I won't even think about what I see there I will look at the reaction itself and I write magnesium and fluoride squared. and I write magnesium and fluoride squared. Now I am ready to plug in what I know. and fluoride squared. Now I am ready to plug in what I know. I know K_sp value was 6.9 x 10 ^ -9 And the magnesium was represented with S and the fluoride, we look right here it is .10 + 2s and now I am ready to square that value. The s is very small. And it is so small, that even if we double it it is going to be very small. And we are going to see that this is true lets assume that term can go away. And that leaves me with 6.9 x 10 ^-9 is equal to 0.10 squared times s. So if we divide both sides by 0.10 squared. We are going to have s equal to 6.9 x 10 -7. Lets verify that this is a good assumption that we made right here. If I double that number and subtract it from 0.10 is it still 0.10? So that was a good assumption that we made. That is the molar solubility of magnesium fluoride in the solution already containing a common ion. Now it might be a good exercise for you to calculate the molar solubility of magnesium fluoride. If this were a zero, in pure water and see if that is bigger if it is bigger then certainly this was and see if that is bigger if it is bigger then certainly this was where you should be. It should decrease the molar solubility of this ion. So this is the end of our learning objective 11. We have learn how to calculate the molar solubility in a solution that contains a common ion. I am going to work several more of these example problems the molar solubility in a solution that contains a common ion. I am going to work several more of these example problems as separate recordings that you may want to watch and work your way through. So that you get really familiar with how to write the reaction and how to solve for molar solubility.

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